Behavior of singularity of $e^{-z^2}$ at infinite

Question

The function $$f(z)=e^{-z^2}$$ doesn't have any singularity in the finite plane. For singularity at the infinite plane $$g(z)=f\left(\frac{1}{z}\right)=e^{-1/z^2}$$ This function has singular behavior at $z=0$. To know what kind of singularity, I have to find the Laurent series. But that's a nontrivial task, For it, we need to find the coefficients: $$c_k=\frac{1}{2\pi i}\oint_C\frac{f(\zeta)}{(\zeta-z_0)^{k+1}} d\zeta$$ I'm not sure, If we can simply use $e^{-z}$ with $z\rightarrow 1/z^2$. Please help me with this.

Answer 1

Just use Taylor series for $e^z = \sum_{n=0} ^\infty \frac{x^n}{n!}$ where $z = - \frac{1}{w^2}$ and it becomes obvious that you have an essential singularity.